NCERT Solutions Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

Q.1 Find the principal values of the following

${\sin }^{–1}\left(–\frac{1}{2}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\sin }^{-1}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{sinx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\sin }^{–1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\sin \left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6}\end{array}$

Q.2

${\cos }^{–1}\left(\frac{\sqrt{3}}{2}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=\frac{\sqrt{3}}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cos }^{-1}\mathrm{is}\\ \left[0,\mathrm{\pi }\right]\mathrm{and}\cos \left(\frac{\mathrm{\pi }}{6}\right)=\frac{\sqrt{3}}{2}.\end{array}$

Q.3

${\text{cosec}}^{–1}\left(2\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}{\mathrm{cosec}}^{-1}\left(2\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosecx}=2.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\mathrm{cosec}}^{-1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]-\left\{0\right\}\mathrm{and}\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)=2.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cosec}}^{-1}\left(2\right)=\frac{\mathrm{\pi }}{6}\end{array}$

Q.4

${\text{tan}}^{–1}(-\sqrt{3})$

Ans.

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}(-\sqrt{3})=\mathrm{x}.\mathrm{Then},\mathrm{tanx}=-\sqrt{3}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\tan }^{–1}\mathrm{is}\\ \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\mathrm{and}\tan \left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\tan }^{-1}(-\sqrt{3})=-\frac{\mathrm{\pi }}{3}.\end{array}$

Q.5

${\text{cos}}^{–1}\left(-\frac{1}{2}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\cos }^{-1}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cos }^{–1}\mathrm{is}\\ [0,\mathrm{\pi }]\mathrm{and}\cos \mathrm{}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}.\\ \mathrm{cosx}=-\frac{1}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\cos \left(\frac{\mathrm{\pi }}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\cos \left(\frac{2\mathrm{\pi }}{3}\right)\\ \Rightarrow \mathrm{x}=\frac{2\mathrm{\pi }}{3}\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{2\mathrm{\pi }}{3}.\end{array}$

Q.6

${\text{tan}}^{–1}(-1)$

Ans.

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\tan }^{-1}(-1)=-1.\\ \mathrm{We}\hspace{0.17em}\hspace{0.17em}\mathrm{know}\hspace{0.17em}\hspace{0.17em}\mathrm{that}\hspace{0.17em}\hspace{0.17em}\mathrm{the}\hspace{0.17em}\hspace{0.17em}\mathrm{range}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}\mathrm{the}\hspace{0.17em}\hspace{0.17em}\mathrm{principle}\hspace{0.17em}\hspace{0.17em}\mathrm{value}\hspace{0.17em}\hspace{0.17em}\mathrm{branch}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\hspace{0.17em}\hspace{0.17em}\\ \left(-\frac{\mathrm{\pi }}{2},\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{\pi }}{2}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\tan \left(\frac{\mathrm{\pi }}{4}\right)\mathrm{\hspace{0.17em}}=1.\\ \mathrm{Therefore},\mathrm{\hspace{0.17em}}\mathrm{principle}\hspace{0.17em}\hspace{0.17em}\mathrm{value}\hspace{0.17em}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}(-1)=-\frac{\mathrm{\pi }}{4}\end{array}$

Q.7

${\text{sec}}^{–1}\left(\frac{2}{\sqrt{3}}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}{\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=\mathrm{x}.\mathrm{Then},\mathrm{secx}=\frac{2}{\sqrt{3}}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cos }^{-1}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\\ [0,\mathrm{\pi }]-\left\{\frac{\mathrm{\pi }}{2}\right\}\mathrm{and}\sec \left(\frac{\mathrm{\pi }}{6}\right)=\frac{2}{\sqrt{3}}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{6}\end{array}$

Q.8

${\text{cot}}^{–1}\left(\sqrt{3}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\cot }^{-1}\left(\sqrt{3}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cotx}=\sqrt{3}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cot }^{–1}\mathrm{is}\\ \left(0,\mathrm{\pi }\right)\mathrm{and}\cot \left(\frac{\mathrm{\pi }}{6}\right)=\sqrt{3}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\cot }^{-1}\left(\sqrt{3}\right)=\frac{\mathrm{\pi }}{6}.\end{array}$

Q.9

${\text{cos}}^{–1}\left(-\frac{1}{\sqrt{2}}\right)$

Ans.

$\begin{array}{l}\mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=-\frac{1}{\sqrt{2}}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cos }^{–1}\mathrm{is}\\ [0,\mathrm{\pi }]\mathrm{and}\cos \left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}.\\ \mathrm{cosx}=-\frac{1}{\sqrt{2}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\cos \left(\frac{\mathrm{\pi }}{4}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\cos \left(\frac{3\mathrm{\pi }}{4}\right)\\ \Rightarrow \mathrm{x}=\frac{3\mathrm{\pi }}{4}\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{3\mathrm{\pi }}{4}.\end{array}$

Q.10

$cose{c}^{–1}(-\sqrt{2})$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}{\mathrm{cosec}}^{-1}(-\sqrt{2})=\mathrm{x}.\mathrm{Then},\mathrm{cosecx}=-\sqrt{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}\\ {\mathrm{cosec}}^{-1}\mathrm{is}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]-\left\{0\right\}\mathrm{and}\mathrm{cosec}\left(\frac{\mathrm{\pi }}{4}\right)=\sqrt{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\mathrm{cosec}}^{-1}(-\sqrt{2})=-\frac{\mathrm{\pi }}{4}\end{array}$

Q.11 Find the values of the following:

${\text{tan}}^{–1}\left(1\right)+{\cos }^{–1}-\left(\frac{1}{2}\right)+{\sin }^{–1}\left(-\frac{1}{2}\right)$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ {\tan }^{–1}\left(\mathrm{1}\right)+{\cos }^{–1}\mathrm{}-\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{\sin }^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{We}\mathrm{shall}\mathrm{find}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{of}\mathrm{each}\mathrm{term}\mathrm{as}\mathrm{given}\mathrm{below}:\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\left(1\right)=\mathrm{x}.\mathrm{Then},\mathrm{tanx}=1.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\tan }^{–1}\mathrm{is}\\ \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\mathrm{and}\mathrm{}\mathrm{t}\mathrm{}\mathrm{an}\left(\frac{\mathrm{\pi }}{4}\right)=1.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}.\\ \mathrm{second}\mathrm{term}:\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\cos }^{-1}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cos }^{–1}\mathrm{is}\\ [0,\mathrm{\pi }]\mathrm{and}\cos \mathrm{}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}.\\ \mathrm{cosx}=-\frac{1}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\cos \left(\frac{\mathrm{\pi }}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\cos \left(\frac{2\mathrm{\pi }}{3}\right)\\ \mathrm{and}\mathrm{third}\mathrm{term}:\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\sin }^{-1}\mathrm{}\left(-\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{sinx}=-\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\sin }^{–1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\sin \left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6}\\ \mathrm{Then},\\ {\tan }^{–1}\left(\mathrm{1}\right)+{\cos }^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)+{\sin }^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{\pi }}{4}+\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\\ \hspace{0.17em}\hspace{0.17em}=\frac{3\mathrm{\pi }+8\mathrm{\pi }-2\mathrm{\pi }}{12}\\ \hspace{0.17em}\hspace{0.17em}=\frac{9\mathrm{\pi }}{12}\\ \hspace{0.17em}\hspace{0.17em}=\frac{3\mathrm{\pi }}{4}.\end{array}$

Q.12

${\text{cos}}^{–1}\left(\frac{1}{2}\right)+2{\sin }^{–1}\left(\frac{1}{2}\right)$

Ans.

$\begin{array}{l}{\cos }^{–1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{}+2{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{First}\mathrm{term}:{\cos }^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{Let}\mathrm{\hspace{0.17em}}{\cos }^{-1}\left(\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{cosx}=\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\cos }^{-1}\mathrm{is}\\ [0,\mathrm{\pi }]\mathrm{and}\cos \left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\cos }^{-1}\mathrm{}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{3}\\ \mathrm{Second}\mathrm{term}:\hspace{0.17em}\hspace{0.17em}2{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\sin }^{-1}\left(\frac{1}{2}\right)=\mathrm{x}.\mathrm{Then},\mathrm{sinx}=\frac{1}{2}.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\sin }^{–1}\mathrm{is}\\ \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\sin \mathrm{}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}.\\ \mathrm{Therefore},\mathrm{}\\ \mathrm{principal}\mathrm{value}\mathrm{of}2{\sin }^{-1}\left(\frac{1}{2}\right)=2\times \frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{3}\\ \therefore \mathrm{\hspace{0.17em}}{\cos }^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+2{\sin }^{–1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{3}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{2\mathrm{\pi }}{3}\end{array}$

Q.13

$\begin{array}{l}{\mathrm{Ifsin}}^{–1}\mathrm{x}=\mathrm{y},\mathrm{then}\\ \\ \mathrm{A})0\le \mathrm{y}\le \mathrm{\pi }\\ \mathrm{B})–\frac{\mathrm{\pi }}{2}\le \mathrm{y}\le \frac{\mathrm{\pi }}{2}\\ \mathrm{C})0<\mathrm{y}<\mathrm{\pi }\\ \mathrm{D})–\frac{\mathrm{\pi }}{2}<\mathrm{y}<\frac{\mathrm{\pi }}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},{\sin }^{-\mathrm{1}}:\mathrm{}[–\mathrm{1},1]\to \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\\ \mathrm{So},\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{.}\end{array}$

Q.14

$\begin{array}{l}{\tan }^{–1}\sqrt{3}–{\sec }^{–1}(–2)\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\mathrm{\pi }\\ \left(\mathrm{B}\right)–\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{C}\right)\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{D}\right)\frac{2\mathrm{\pi }}{3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{–1}\sqrt{\mathrm{3}}-{\sec }^{–1}(-\mathrm{2})\\ \mathrm{Since},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\tan }^{–1}\sqrt{\mathrm{3}}=\frac{\mathrm{\pi }}{3},\mathrm{and}\\ \mathrm{let}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\sec }^{-1}(-2)=\mathrm{x}.\mathrm{Then},\mathrm{secx}=-\mathrm{\hspace{0.17em}}2.\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{principal}\mathrm{value}\mathrm{branch}\mathrm{of}{\sec }^{-1}\mathrm{is}\\ [0,\mathrm{\pi }]-\left\{\frac{\mathrm{\pi }}{2}\right\}\mathrm{and}\sec \left(\frac{\mathrm{\pi }}{3}\right)=2.\\ \therefore \mathrm{secx}=-2=-\sec \left(\frac{\mathrm{\pi }}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\sec \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\sec \left(\frac{2\mathrm{\pi }}{3}\right)\\ \mathrm{Therefore},\mathrm{principal}\mathrm{value}\mathrm{of}{\sec }^{-1}(-2)=\frac{2\mathrm{\pi }}{3}.\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\tan }^{–1}\sqrt{\mathrm{3}}-{\sec }^{–1}(-\mathrm{2})=\frac{\mathrm{\pi }}{3}-\left(\frac{2\mathrm{\pi }}{3}\right)\\ =\frac{\mathrm{\pi }-2\mathrm{\pi }}{3}\\ =-\frac{\mathrm{\pi }}{3}.\\ \mathrm{Option}\hspace{0.17em}\hspace{0.17em}\mathrm{B}\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}.\end{array}$

Q.15 Prove the following:

${\text{3sin}}^{–1}\mathrm{x}={\sin }^{–1}(3\mathrm{x}–4{\mathrm{x}}^3),\mathrm{x}\in \left[-\frac{1}{2},\frac{1}{2}\right]$

Ans.

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.={\sin }^{-1}(3\mathrm{x}-4{\mathrm{x}}^3)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\sin }^{-1}(3\mathrm{sin\theta }-4{\sin }^3\mathrm{\theta })[\mathrm{Putting}\mathrm{x}=\mathrm{sin\theta }]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\sin }^{-1}\left(\sin 3\mathrm{\theta }\right)[\because \sin 3\mathrm{x}=3\mathrm{sinx}-4{\sin }^3\mathrm{x}]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=3\mathrm{\theta }\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=3{\sin }^{-1}\mathrm{x}\left[\therefore \mathrm{x}=\mathrm{sin\theta }\Rightarrow \mathrm{\theta }={\sin }^{-1}\mathrm{x}\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.16

${\text{3cos}}^{–1}x=co{s}^{–1}\left(4x^3-3x\right),x\in \left[\frac{1}{2},1\right]$

Ans.

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.={\cos }^{-1}(4{\mathrm{x}}^3-3\mathrm{x})\\ ={\cos }^{-1}(4{\cos }^3\mathrm{\theta }-3\mathrm{cos\theta })[\mathrm{Putting}\mathrm{x}=\mathrm{cos\theta }]\\ ={\cos }^{-1}\left(\cos 3\mathrm{\theta }\right)[\because \cos 3\mathrm{x}=4{\cos }^3\mathrm{x}-3\mathrm{cosx}]\\ =3\mathrm{\theta }\\ =3{\cos }^{-1}\mathrm{x}\left[\therefore \mathrm{x}=\mathrm{cos\theta }\Rightarrow \mathrm{\theta }={\cos }^{-1}\mathrm{x}\right]\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.17

${\text{tan}}^{–1}\left(\frac{2}{11}\right)+{\tan }^{–1}\left(\frac{7}{24}\right)={\tan }^{–1}\left(\frac{1}{2}\right)$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.={\tan }^{–1}\left(\frac{\mathrm{2}}{\mathrm{11}}\right)+{\tan }^{–1}\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{–1}\mathrm{}\left(\frac{\frac{\mathrm{2}}{\mathrm{11}}\mathrm{+}\frac{\mathrm{7}}{\mathrm{24}}}{1-\frac{2}{11}\times \frac{7}{24}}\right)\mathrm{}\left[\because {\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left(\frac{\frac{48+77}{11\times 24}}{\frac{11\times 24-14}{11\times 24}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left(\frac{125}{250}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left(\frac{1}{2}\right)\hspace{0.17em}\hspace{0.17em}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.18

${\text{2tan}}^{–1}\left(\frac{1}{2}\right)+{\tan }^{–1}\left(\frac{1}{7}\right)={\tan }^{–1}\left(\frac{31}{17}\right)$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=2{\tan }^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{\tan }^{–1}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\left(\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}\right)+{\tan }^{–1}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\left[\because 2{\tan }^{-1}\mathrm{x}={\tan }^{-1}\left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right)\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\left(\frac{4}{3}\right)+{\tan }^{–1}\mathrm{}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)\left[\because {\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\mathrm{}\left(\frac{\mathrm{31}}{\mathrm{17}}\right)\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.19 Write the following functions in the simplest form:

${\text{tan}}^{–1}\frac{\sqrt{1+{\mathrm{x}}^2}–1}{\mathrm{x}},\mathrm{x}\ne 0$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ {\tan }^{-1}\frac{\sqrt{1+{\mathrm{x}}^2}-1}{\mathrm{x}}={\tan }^{-1}\frac{\sqrt{1+{\tan }^2\mathrm{\theta }}-1}{\mathrm{tan\theta }}\mathrm{\hspace{0.17em}}[\mathrm{Putting}\mathrm{x}=\mathrm{tan\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\frac{\sqrt{{\sec }^2\mathrm{\theta }}-1}{\mathrm{tan\theta }}[\because 1+{\tan }^2\mathrm{\theta }={\sec }^2\mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\frac{\mathrm{sec\theta }-1}{\mathrm{tan\theta }}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{-1}\left(\tan \left(\frac{\mathrm{\theta }}{2}\right)\right)[\because {\tan }^{-1}\left(\mathrm{tanx}\right)=\mathrm{x}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{\theta }}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{2}{\tan }^{-1}\mathrm{x}\end{array}$

Q.20

${\text{tan}}^{–1}\frac{1}{\sqrt{{\mathrm{x}}^2-1}},\left|\mathrm{x}\right|>1$

Ans.

${\text{tan}}^{–1}\frac{1}{\sqrt{{\mathrm{x}}^2-1}},\left|\mathrm{x}\right|>1$

Q.21

ta{n}^{–1}\left(\sqrt{\frac{1–cosx}{1+cosx}}\right),x<\pi

Ans.

\begin{array}{l}{\text{tan}}^{\text{-1}}\left(\sqrt{\frac{\text{1-cosx}}{\text{1+cosx}}}\right)={\text{tan}}^{\text{-1}}\left(\sqrt{\frac{{\text{2sin}}^{\text{2}}\left(\frac{\text{x}}{2}\right)}{{\text{2cos}}^{\text{2}}\left(\frac{x}{2}\right)}}\right)\left[\begin{array}{l}\because \cos 2x=1-2{\sin }^{2}x\\ =2{\cos }^{2}x-1\end{array}\right]\\ ={\tan }^{-1}\left(\tan \frac{x}{2}\right)\\ =\frac{x}{2}\left[\because {\tan }^{-1}\left(\tan x\right)=x\right]\end{array}

Q.22

${\text{tan}}^{–1}\left(\frac{\mathrm{cosx}–\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\right),0<\mathrm{x}<\mathrm{\pi }$

Ans.

$\begin{array}{l}{\tan }^{–1}\left(\frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}}\right)={\tan }^{-1}\left(\frac{1-\mathrm{tanx}}{1+\mathrm{tanx}}\right)\left[\begin{array}{l}\mathrm{Dividing}\mathrm{numerator}\\ \mathrm{and}\mathrm{denominator}\mathrm{by}\mathrm{sinx}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\right\}\left[\because \tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=\frac{1-\mathrm{tanx}}{1+\mathrm{tanx}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\pi }}{4}-\mathrm{x}[\because {\tan }^{-1}\left(\mathrm{tanx}\right)=\mathrm{x}]\end{array}$

Q.23

${\text{tan}}^{–1}\frac{\mathrm{x}}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}},\left|\mathrm{x}\right|<\mathrm{a}$

Ans.

$\begin{array}{l}{\tan }^{–1}\frac{\mathrm{x}}{\sqrt{{\mathrm{a}}^{\mathrm{2}}–{\mathrm{x}}^{\mathrm{2}}}}={\tan }^{–1}\frac{\mathrm{asin\theta }}{\sqrt{{\mathrm{a}}^{\mathrm{2}}-{\mathrm{a}}^{\mathrm{2}}{\sin }^2\mathrm{\theta }}}[\mathrm{Putting}\mathrm{x}=\mathrm{a}\sin \mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\mathrm{}\left(\frac{\mathrm{asin\theta }}{\mathrm{acos\theta }}\right)[\because 1-{\sin }^2\mathrm{\theta }={\cos }^2\mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\mathrm{tan\theta }\right)\left[\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\mathrm{tan\theta }\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{\theta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)[\because \mathrm{x}=\mathrm{asin\theta }]\end{array}$

Q.24

${\text{tan}}^{–1}\left(\frac{3{\mathrm{a}}^2\mathrm{x}–{\mathrm{x}}^3}{{\mathrm{a}}^3–3{\mathrm{ax}}^2}\right),\mathrm{a}>0;-\frac{\mathrm{a}}{\sqrt{3}}\le \mathrm{x}\le \frac{\mathrm{a}}{\sqrt{3}}$

Ans.

$\begin{array}{l}{\tan }^{–1}\mathrm{}\left(\frac{3{\mathrm{a}}^{\mathrm{2}}\mathrm{x}-{\mathrm{x}}^{\mathrm{3}}}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{ax}}^{\mathrm{2}}}\right)={\tan }^{–1}\mathrm{}\left(\frac{3{\mathrm{a}}^{\mathrm{3}}\mathrm{tan\theta }-{\left(\mathrm{atan\theta }\right)}^{\mathrm{3}}}{{\mathrm{a}}^{\mathrm{3}}-3\mathrm{a}{\left(\mathrm{atan\theta }\right)}^{\mathrm{2}}}\right)[\mathrm{Putting}\mathrm{x}=\mathrm{atan\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{–1}\left(\frac{3{\mathrm{a}}^{\mathrm{3}}\mathrm{tan\theta }-{\mathrm{a}}^3{\tan }^3\mathrm{\theta }}{{\mathrm{a}}^{\mathrm{3}}-3{\mathrm{a}}^{\mathrm{3}}{\tan }^2\mathrm{\theta }}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{–1}\mathrm{}\left\{\frac{\bcancel{{\mathrm{a}}^3}(\mathrm{3}\mathrm{tan\theta }-{\tan }^3\mathrm{\theta })}{\bcancel{{\mathrm{a}}^{\mathrm{3}}}(\mathrm{1}-\mathrm{3}{\tan }^2\mathrm{\theta })}\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{–1}\left(\tan 3\mathrm{\theta }\right)\left[\tan 3\mathrm{\theta }=\frac{3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }}{1-3{\tan }^2\mathrm{\theta }}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3\mathrm{\theta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)[\because \mathrm{x}=\mathrm{atan\theta }]\end{array}$

Q.25 Find the values of each of the following:

${\text{tan}}^{–1}\left\{2\cos \left(2{\sin }^{–1}\frac{1}{2}\right)\right\}$

Ans.

$\begin{array}{l}{\tan }^{-1}\left\{2\cos \left(2{\sin }^{–1}\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}{\tan }^{-1}\mathrm{}\left\{2\cos \left(\mathrm{2}\times \frac{\mathrm{\pi }}{6}\right)\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left\{2\cos \left(\frac{\mathrm{\pi }}{3}\right)\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left\{\mathrm{2}\times \frac{1}{2}\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left(1\right)\mathrm{\hspace{0.17em}}=\frac{\mathrm{\pi }}{4}\end{array}$

Q.26

$\text{cot}\left({\tan }^{–1}\mathrm{a}+{\cot }^{–1}\mathrm{a}\right)$

Ans.

$\begin{array}{l}\cot \mathrm{}\left({\tan }^{–1}\mathrm{a}+{\cot }^{–1}\mathrm{a}\right)=\cot \left[\because {\tan }^{–1}\mathrm{x}+{\cot }^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\mathrm{}\left(\frac{\mathrm{\pi }}{2}\right)\mathrm{}\\ =0\end{array}$

Q.27

$\text{tan}\frac{1}{2}\left({\sin }^{–1}\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}+{\cos }^{–1}\frac{1–{\mathrm{y}}^2}{1+{\mathrm{y}}^2}\right),\left|\mathrm{x}\right|<1,\mathrm{y}>0\mathrm{and}\mathrm{xy}<1$

Ans.

$\begin{array}{l}\tan \frac{\mathrm{1}}{\mathrm{2}}\mathrm{}\left({\sin }^{–1}\frac{2\mathrm{x}}{\mathrm{1}+{\mathrm{x}}^{\mathrm{2}}}+{\cos }^{–1}\frac{\mathrm{1}-{\mathrm{y}}^{\mathrm{2}}}{\mathrm{1}+{\mathrm{y}}^{\mathrm{2}}}\right)\\ =\tan \frac{\mathrm{1}}{\mathrm{2}}\mathrm{}\left({\sin }^{–1}\frac{\mathrm{2}\mathrm{tan\alpha }}{\mathrm{1}+{\tan }^2\mathrm{\alpha }}+{\cos }^{–1}\frac{\mathrm{1}-{\tan }^2\mathrm{\beta }}{\mathrm{1}+{\tan }^2\mathrm{\beta }}\right)[\mathrm{Putting}\mathrm{x}=\mathrm{tan\alpha },\mathrm{}\mathrm{y}=\mathrm{tan\beta }]\\ =\tan \frac{\mathrm{1}}{\mathrm{2}}\mathrm{}\left\{{\sin }^{–1}\left(\sin 2\mathrm{\alpha }\right)+{\cos }^{–1}\left(\cos 2\mathrm{\beta }\right)\right\}\\ =\tan \frac{\mathrm{1}}{\mathrm{2}}(2\mathrm{\alpha }+2\mathrm{\beta })\\ =\tan (\mathrm{\alpha }+\mathrm{\beta })\\ =\frac{\mathrm{tan\alpha }+\mathrm{tan\beta }}{1-\mathrm{tan\alpha tan\beta }}\\ =\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\end{array}$

Q.28

$\mathrm{If}\sin \left({\sin }^{–1}\frac{1}{5}+{\cos }^{–1}\mathrm{x}\right)=1,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{.}$

Ans.

$\begin{array}{l}\sin \left({\sin }^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\cos }^{–1}\mathrm{x}\right)=\mathrm{1}\\ \Rightarrow \sin \left({\sin }^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\cos }^{–1}\mathrm{x}\right)\mathrm{}={\sin }^{–1}\left(\frac{\mathrm{\pi }}{2}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\sin }^{–1}\frac{\mathrm{1}}{\mathrm{5}}+{\cos }^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\sin }^{–1}\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{\pi }}{2}-{\cos }^{–1}\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\sin }^{–1}\frac{\mathrm{1}}{\mathrm{5}}={\sin }^{–1}\mathrm{x}\left[\because {\sin }^{–1}\mathrm{x}+{\cos }^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{5}=\mathrm{x}\end{array}$

Q.29

${\text{If tan}}^{–1}\frac{\mathrm{x}–1}{\mathrm{x}–2}+{\tan }^{–1}\frac{\mathrm{x}+1}{\mathrm{x}+2}=\frac{\mathrm{\pi }}{4},\mathrm{}\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}.$

Ans.

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\tan }^{–1}\frac{\mathrm{x}-\mathrm{1}}{\mathrm{x}-\mathrm{2}}+{\tan }^{–1}\frac{\mathrm{x}+1}{\mathrm{x}+2}=\frac{\mathrm{\pi }}{4}\\ \Rightarrow {\tan }^{–1}\left(\frac{\frac{\mathrm{x}-\mathrm{1}}{\mathrm{x}-\mathrm{2}}+\frac{\mathrm{x}+1}{\mathrm{x}+2}}{1-\left(\frac{\mathrm{x}-\mathrm{1}}{\mathrm{x}-\mathrm{2}}\right)\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)}\right)=\frac{\mathrm{\pi }}{4}\left[\because {\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\frac{\frac{{\mathrm{x}}^{\mathrm{2}}+\mathrm{x}-2+{\mathrm{x}}^{\mathrm{2}}-\mathrm{x}-\mathrm{2}}{(\mathrm{x}-2)(\mathrm{x}+2)}}{\frac{{\mathrm{x}}^{\mathrm{2}}-\mathrm{4}-{\mathrm{x}}^{\mathrm{2}}+1}{(\mathrm{x}-2)(\mathrm{x}+2)}}\right)=\tan \frac{\mathrm{\pi }}{4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\frac{2{\mathrm{x}}^2-4}{-3}\right)=1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}2{\mathrm{x}}^2-4=-3\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\pm \frac{1}{\sqrt{2}}\end{array}$

Q.30

${\text{sin}}^{–1}\left(\sin \frac{2\mathrm{\pi }}{3}\right)$

Ans.

$\begin{array}{l}\mathrm{\hspace{0.17em}}{\sin }^{–1}\mathrm{}\left(\sin \frac{\mathrm{2}\mathrm{\pi }}{\mathrm{3}}\right)\mathrm{}=\sin \because \mathrm{Range}\mathrm{of}\mathrm{sin\theta }\mathrm{is}\hspace{0.17em}{\left[-\frac{\mathrm{\pi }}{2}\mathrm{,}\frac{\mathrm{\pi }}{2}\mathrm{}\right]}^{–1}\mathrm{}\left\{\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{\mathrm{3}}\right)\right\}\\ ={\sin }^{–1}\left\{\sin \frac{\mathrm{\pi }}{\mathrm{3}}\right\}\left[\because \sin (\mathrm{\pi }-\mathrm{\theta })=\mathrm{sin\theta }\right]\\ =\frac{\mathrm{\pi }}{\mathrm{3}}\end{array}$

Q.31

${\text{tan}}^{–1}\left(tan\frac{3\pi }{4}\right)$

Ans.

$\begin{array}{l}{\tan }^{–1}\left(\tan \frac{\mathrm{3}\mathrm{\pi }}{\mathrm{4}}\right)={\tan }^{–1}\left\{\tan \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{\mathrm{4}}\right)\right\}\left[\because \mathrm{Range}\mathrm{of}\mathrm{tan\theta }\mathrm{is}\hspace{0.17em}\left(-\frac{\mathrm{\pi }}{2}\mathrm{,}\frac{\mathrm{\pi }}{2}\right)\right]\\ ={\tan }^{–1}\mathrm{}\left\{-\tan \frac{\mathrm{\pi }}{\mathrm{4}}\right\}[\because \tan (\mathrm{\pi }-\mathrm{\theta })=-\mathrm{tan\theta }]\\ ={\tan }^{–1}\left\{\tan \left(-\frac{\mathrm{\pi }}{\mathrm{4}}\right)\right\}\mathrm{}[\tan (-\mathrm{\theta })=-\mathrm{tan\theta }]\\ =-\frac{\mathrm{\pi }}{\mathrm{4}}\end{array}$

Q.32

$\text{tan}\left({\sin }^{–1}\frac{3}{5}+{\cot }^{–1}\frac{3}{2}\right)$

Ans.

$\begin{array}{l}\tan \mathrm{}\left({\sin }^{–1}\frac{\mathrm{3}}{\mathrm{5}}+{\cot }^{–1}\frac{\mathrm{3}}{\mathrm{2}}\right)\\ \mathrm{Let}{\sin }^{–1}\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{x}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}{\cot }^{–1}\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{y}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{sinx}=\frac{3}{5}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{coty}=\frac{3}{2}\\ \Rightarrow \mathrm{cosx}=\sqrt{1-{\sin }^2\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\frac{4}{5}\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\tan \mathrm{}\mathrm{y}=\frac{1}{\cot \mathrm{y}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{\left(\frac{3}{2}\right)}=\frac{2}{3}\\ \Rightarrow \mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}=\frac{\left(\frac{3}{5}\right)}{\left(\frac{4}{5}\right)}\\ \hspace{0.17em}\hspace{0.17em}=\frac{3}{4}\\ \Rightarrow \mathrm{x}={\tan }^{-1}\left(\frac{3}{4}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{y}={\tan }^{-1}\left(\frac{2}{3}\right)\\ \Rightarrow {\sin }^{–1}\frac{\mathrm{3}}{\mathrm{5}}={\tan }^{-1}\left(\frac{3}{4}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}{\cot }^{–1}\frac{\mathrm{3}}{\mathrm{2}}={\tan }^{-1}\left(\frac{2}{3}\right)\\ \therefore \tan \left({\sin }^{–1}\frac{\mathrm{3}}{\mathrm{5}}+{\cot }^{–1}\frac{\mathrm{3}}{\mathrm{2}}\right)=\tan \left\{{\tan }^{-1}\mathrm{}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right\}\\ =\tan \left\{{\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)\right\}\\ =\tan \left\{{\tan }^{-1}\left(\frac{17}{6}\right)\right\}\\ =\frac{17}{6}\end{array}$

Q.33

$\begin{array}{l}{\cos }^{–1}\left(\cos \frac{7\mathrm{\pi }}{6}\right)\mathrm{}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\frac{7\mathrm{\pi }}{6}\\ \left(\mathrm{B}\right)\frac{5\mathrm{\pi }}{6}\\ \left(\mathrm{C}\right)\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{D}\right)\frac{\mathrm{\pi }}{6}\end{array}$

Ans.

$\begin{array}{l}{\cos }^{–1}\mathrm{}\left(\cos \frac{\mathrm{7}\mathrm{\pi }}{\mathrm{6}}\right)={\cos }^{–1}\left\{\cos \left(2\mathrm{\pi }-\frac{5\mathrm{\pi }}{\mathrm{6}}\right)\right\}[\because \mathrm{Range}\mathrm{of}\mathrm{cos\theta }\mathrm{is}\mathrm{\hspace{0.17em}}\left[0\mathrm{,}\mathrm{\pi }\right]]\\ ={\cos }^{–1}\mathrm{}\left\{\cos \left(\frac{5\mathrm{\pi }}{\mathrm{6}}\right)\right\}[\because \cos (2\mathrm{\pi }-\mathrm{\theta })=\mathrm{cos\theta }]\\ =\frac{5\mathrm{\pi }}{\mathrm{6}}[\because {\cos }^{-1}\left(\mathrm{cosx}\right)=\mathrm{x}]\\ \mathrm{Hence}\mathrm{option}\left(\mathrm{b}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.34

$\begin{array}{l}\sin \left(\frac{\mathrm{\pi }}{3}-{\sin }^{–1}\left(-\frac{1}{2}\right)\right)\mathrm{}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\frac{1}{2}\\ \left(\mathrm{B}\right)\frac{1}{3}\\ \left(\mathrm{C}\right)\frac{1}{4}\\ \left(\mathrm{D}\right)1\end{array}$

Ans.

$\begin{array}{l}\sin \left(\frac{\mathrm{\pi }}{\mathrm{3}}-{\sin }^{–1}\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=\sin \mathrm{}\left(\frac{\mathrm{\pi }}{\mathrm{3}}+{\sin }^{–1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)[\sin (-\mathrm{\theta })=-\mathrm{sin\theta }]\\ =\sin \mathrm{}\left(\frac{\mathrm{\pi }}{\mathrm{3}}+\frac{\mathrm{\pi }}{\mathrm{6}}\right)\left[\because \sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}\right]\\ =\sin \mathrm{}\left(\frac{2\mathrm{\pi }+\mathrm{\pi }}{\mathrm{6}}\right)\\ =\sin \left(\mathrm{}\frac{3\mathrm{\pi }}{\mathrm{6}}\right)\\ =\sin \mathrm{}\left(\frac{\mathrm{\pi }}{\mathrm{2}}\right)\\ =1\left[\because \sin \frac{\mathrm{\pi }}{2}=1\right]\\ \mathrm{Hence},\mathrm{option}\left(\mathrm{D}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.35

$\begin{array}{l}{\tan }^{-1}\sqrt{\mathrm{3}}-{\cot }^{-1}\left(\mathrm{–}\sqrt{\mathrm{3}}\right)\mathrm{is}\mathrm{\hspace{0.33em}}\mathrm{equal}\mathrm{\hspace{0.33em}}\mathrm{to}\\ \begin{array}{l}\left(\mathrm{A}\right)\hspace{0.33em}\mathrm{\pi }\\ \left(\mathrm{B}\right)\mathrm{\hspace{0.33em}}-\frac{\mathrm{\pi }}{\mathrm{2}}\\ \left(\mathrm{C}\right)\mathrm{\hspace{0.33em}}0\\ \left(\mathrm{D}\right)\mathrm{\hspace{0.33em}}2\sqrt{\mathrm{3}}\end{array}\end{array}$

Ans.

$\begin{array}{l}{\tan }^{–1}\sqrt{\mathrm{3}}-{\cot }^{–1}(-\sqrt{\mathrm{3}})\\ =-\mathrm{\hspace{0.17em}}{\tan }^{–1}(-\sqrt{\mathrm{3}})-{\cot }^{–1}(-\sqrt{\mathrm{3}})[\because \tan (-\mathrm{\theta })=\mathrm{tan\theta }]\\ =-\mathrm{\hspace{0.17em}}\{{\tan }^{–1}(-\sqrt{\mathrm{3}})+{\cot }^{–1}(-\sqrt{\mathrm{3}})\}\\ =-\frac{\mathrm{\pi }}{2}[\because {\tan }^{-1}\mathrm{x}+{\mathrm{con}}^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2}]\\ \mathrm{Hence}\mathrm{option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.36

${\text{Find the value of cos}}^{–1}\left(\cos \frac{13\mathrm{\pi }}{6}\right).$

Ans.

$\begin{array}{l}\mathrm{Range}\mathrm{of}\mathrm{cos\theta }\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{\hspace{0.17em}}[0,\mathrm{\pi }]\\ \therefore {\cos }^{-1}\left(\cos \frac{13\mathrm{\pi }}{6}\right)={\cos }^{-1}\left\{\cos \left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\cos }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{6}\right)\right\}[\because \cos (2\mathrm{\pi }+\mathrm{\theta })=\cos \mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\pi }}{6}\end{array}$

Q.37

${\text{Find the value of tan}}^{–1}\left(\tan \frac{7\mathrm{\pi }}{6}\right).$

Ans.

$\begin{array}{l}\mathrm{Range}\mathrm{of}\mathrm{tan\theta }\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{\hspace{0.17em}}\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\\ \therefore {\tan }^{-1}\left(\tan \frac{7\mathrm{\pi }}{6}\right)={\tan }^{-1}\left\{\tan \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left\{\left(\tan \frac{\mathrm{\pi }}{6}\right)\right\}[\because \tan (\mathrm{\pi }+\mathrm{\theta })=\mathrm{tan\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\pi }}{6}\end{array}$

Q.38

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ 2{\sin }^{–1}\left(\frac{3}{5}\right)={\tan }^{–1}\frac{24}{7}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}{\sin }^{–1}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)=\mathrm{x}\Rightarrow \mathrm{sinx}=\frac{3}{5}\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{cosx}=\sqrt{1-{\sin }^2\mathrm{x}}\\ =\sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ =\frac{4}{5}\\ \hspace{0.17em}\mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}\\ =\frac{\left(\frac{3}{5}\right)}{\left(\frac{4}{5}\right)}\\ =\frac{3}{4}\\ \mathrm{Since},\\ \tan 2\mathrm{x}=\frac{2\tan \mathrm{x}}{1+{\tan }^2\mathrm{x}}\\ =\frac{2\times \frac{3}{4}}{1-{\left(\frac{3}{4}\right)}^2}\\ =\frac{24}{17}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{x}={\tan }^{-1}\left(\frac{24}{17}\right)\\ 2{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)={\tan }^{-1}\left(\frac{24}{17}\right)\end{array}$

Q.39

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\sin }^{–1}\left(\frac{8}{17}\right)+{\sin }^{–1}\left(\frac{3}{5}\right)={\tan }^{–1}\frac{77}{36}\end{array}$

Ans.

$\begin{array}{l}{\sin }^{–1}\left(\frac{\mathrm{8}}{\mathrm{17}}\right)\mathrm{}+{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)={\tan }^{–1}\frac{\mathrm{77}}{\mathrm{36}}\\ \mathrm{Let}{\sin }^{–1}\left(\frac{\mathrm{8}}{\mathrm{17}}\right)=\mathrm{x}\Rightarrow \mathrm{sinx}=\frac{8}{17}\\ \mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{cosx}=\sqrt{1-{\sin }^2\mathrm{x}}\\ =\sqrt{1-{\left(\frac{8}{17}\right)}^2}\\ =\frac{15}{17}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}\\ =\frac{\left(\frac{8}{17}\right)}{\left(\frac{15}{17}\right)}\\ =\frac{8}{15}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}={\tan }^{-1}\frac{8}{15}\\ \Rightarrow {\sin }^{–1}\mathrm{}\left(\frac{\mathrm{8}}{\mathrm{17}}\right)={\tan }^{-1}\frac{8}{15}\\ \mathrm{Again},\mathrm{\hspace{0.17em}}\mathrm{let}\hspace{0.17em}\hspace{0.17em}{\sin }^{–1}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)=\mathrm{y}\Rightarrow \mathrm{siny}=\frac{3}{5}\\ \mathrm{so},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{tany}=\frac{\mathrm{siny}}{\sqrt{1-{\sin }^2\mathrm{y}}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\left(\frac{3}{5}\right)}{\sqrt{1-{\left(\frac{3}{5}\right)}^2}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{3}{4}\\ \mathrm{y}={\tan }^{-1}\left(\frac{3}{4}\right)\\ \therefore \hspace{0.17em}\hspace{0.17em}{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)={\tan }^{-1}\left(\frac{3}{4}\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}={\sin }^{–1}\mathrm{}\left(\frac{\mathrm{8}}{\mathrm{17}}\right)\mathrm{+}\mathrm{si}{\mathrm{n}}^{–1}\mathrm{}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)\\ ={\tan }^{-1}\left(\frac{8}{15}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)\\ ={\tan }^{-1}\left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}\right)\\ ={\tan }^{-1}\left(\frac{77}{36}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.40

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\cos }^{–1}\left(\frac{4}{5}\right)+{\cos }^{–1}\left(\frac{12}{13}\right)={\cos }^{–1}\frac{33}{65}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}{\cos }^{–1}\mathrm{}\left(\frac{4}{5}\right)=\mathrm{x}\mathrm{and}{\cos }^{–1}\mathrm{}\left(\frac{12}{13}\right)=\mathrm{y}\\ \Rightarrow \mathrm{cosx}=\frac{4}{5}\mathrm{and}\mathrm{cosy}=\frac{12}{13}\\ \Rightarrow \mathrm{sinx}=\sqrt{1-{\cos }^2\mathrm{x}}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{siny}=\sqrt{1-{\cos }^2\mathrm{y}}\\ \Rightarrow \mathrm{sinx}=\sqrt{1-{\left(\frac{4}{5}\right)}^2}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{siny}=\sqrt{1-{\left(\frac{12}{13}\right)}^2}\\ \Rightarrow \mathrm{sinx}=\frac{3}{5}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{siny}=\frac{5}{13}\\ \mathrm{Then},\\ \cos (\mathrm{x}+\mathrm{y})=\mathrm{cosx}\mathrm{cosy}-\mathrm{sinx}\mathrm{siny}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{4}{5}\times \frac{12}{13}-\frac{3}{5}\times \frac{5}{13}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{48-15}{5\times 13}\\ \Rightarrow \mathrm{x}+\mathrm{y}={\cos }^{-1}\left(\frac{33}{65}\right)\\ \Rightarrow {\cos }^{–1}\mathrm{}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)+\hspace{0.17em}\mathrm{\hspace{0.17em}}{\cos }^{–1}\mathrm{}\left(\frac{\mathrm{12}}{\mathrm{13}}\right)={\cos }^{–1}\mathrm{}\frac{\mathrm{33}}{\mathrm{65}}.\end{array}$

Q.41

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\cos }^{–1}\left(\frac{12}{13}\right)+{\sin }^{–1}\left(\frac{3}{5}\right)={\sin }^{–1}\frac{56}{65}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}{\cos }^{–1}\mathrm{}\left(\frac{\mathrm{12}}{\mathrm{13}}\right)=\mathrm{x}\mathrm{and}{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)=\mathrm{y}\\ \Rightarrow \mathrm{cosx}=\frac{\mathrm{12}}{\mathrm{13}}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{siny}=\frac{\mathrm{3}}{\mathrm{5}}\\ \Rightarrow \mathrm{sinx}=\sqrt{1-{\cos }^2\mathrm{x}}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{cosy}=\sqrt{1-{\sin }^2\mathrm{y}}\\ \Rightarrow \mathrm{sinx}=\sqrt{1-{\left(\frac{12}{13}\right)}^2}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{cosy}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ \Rightarrow \mathrm{sinx}=\frac{5}{13}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{cosy}=\frac{4}{5}\\ \mathrm{Since},\\ \mathrm{\hspace{0.17em}}\sin (\mathrm{x}+\mathrm{y})=\mathrm{sinx}\mathrm{cosy}+\mathrm{siny}\mathrm{cosx}\\ \hspace{0.17em}\hspace{0.17em}=\left(\frac{5}{13}\right)\left(\frac{4}{5}\right)+\left(\frac{3}{5}\right)\left(\frac{12}{13}\right)\\ \hspace{0.17em}\hspace{0.17em}=\frac{20+36}{65}\\ \mathrm{\hspace{0.17em}}=\frac{56}{65}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{y}={\sin }^{-1}\left(\frac{56}{65}\right)\\ \Rightarrow {\cos }^{–1}\left(\frac{\mathrm{12}}{\mathrm{13}}\right)+{\sin }^{–1}\mathrm{}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)={\sin }^{-1}\left(\frac{56}{65}\right).\end{array}$

Q.42

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\tan }^{–1}\left(\frac{63}{16}\right)={\sin }^{–1}\left(\frac{5}{13}\right)+{\cos }^{–1}\frac{3}{5}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}{\sin }^{–1}\left(\frac{\mathrm{5}}{\mathrm{13}}\right)=\mathrm{x}\mathrm{and}{\cos }^{–1}\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{y}\\ \Rightarrow \mathrm{sinx}=\frac{5}{13}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{cosy}=\frac{3}{5}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{cosx}=\sqrt{1-{\sin }^2\mathrm{x}}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{siny}=\sqrt{1-{\cos }^2\mathrm{y}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{cosx}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{siny}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{cosx}=\frac{12}{13}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{siny}=\frac{4}{5}\\ \Rightarrow \mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}=\frac{5}{12}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{tany}=\frac{4}{3}\\ \mathrm{Since}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\tan (\mathrm{x}+\mathrm{y})=\frac{\mathrm{tanx}+\mathrm{tany}}{1-\mathrm{tanxtany}}\\ =\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times \frac{4}{3}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{y}={\tan }^{-1}\left(\frac{63}{16}\right)\\ \Rightarrow {\sin }^{–1}\mathrm{}\left(\frac{\mathrm{5}}{\mathrm{13}}\right)+{\cos }^{–1}\frac{\mathrm{3}}{\mathrm{5}}={\tan }^{-1}\left(\frac{63}{16}\right)\end{array}$

Q.43

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\tan }^{–1}\left(\frac{1}{5}\right)+{\tan }^{–1}\left(\frac{1}{7}\right)+{\tan }^{–1}\left(\frac{1}{3}\right)+{\tan }^{–1}\left(\frac{1}{8}\right)=\frac{\mathrm{\pi }}{4}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{\hspace{0.17em}}{\tan }^{–1}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)+{\tan }^{–1}\left(\frac{\mathrm{1}}{\mathrm{7}}\right)+{\tan }^{–1}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+{\tan }^{–1}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}\times \frac{1}{7}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}\times \frac{1}{8}}\right)\\ [{\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{12}{34}\right)+{\tan }^{-1}\left(\frac{11}{23}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{–1}\left(\frac{\frac{12}{34}+\frac{11}{23}}{1-\frac{12}{34}\times \frac{11}{23}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left(\frac{650}{650}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\tan }^{-1}\left(1\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\pi }}{4}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.44

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\tan }^{–1}\sqrt{\mathrm{x}}=\frac{1}{2}{\cos }^{–1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right),\hspace{0.17em}\hspace{0.17em}\mathrm{x}\in [0,1]\end{array}$

Ans.

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{1}}{\mathrm{2}}{\cos }^{–1}\left(\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{1}}{\mathrm{2}}{\cos }^{–1}\left(\frac{\mathrm{1}-{\tan }^{\mathrm{2}}\mathrm{\theta }}{1+{\tan }^{\mathrm{2}}\mathrm{\theta }}\right)[\mathrm{Putting}\mathrm{x}={\tan }^{\mathrm{2}}\mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{1}}{\mathrm{2}}{\cos }^{–1}\left(\cos 2\mathrm{\theta }\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2\mathrm{\theta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2{\tan }^{-1}\sqrt{\mathrm{x}}\left[\mathrm{since},\mathrm{\hspace{0.17em}}\mathrm{tan\theta }\mathrm{=}\sqrt{\mathrm{x}}\Rightarrow \mathrm{\theta }={\tan }^{-1}\sqrt{\mathrm{x}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{L}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.45

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\cot }^{–1}\left(\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1–\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}–\sqrt{1–\mathrm{sinx}}}\right)=\frac{\mathrm{x}}{2},\hspace{0.17em}\hspace{0.17em}\mathrm{x}\in (0,\frac{\mathrm{\pi }}{4})\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ ={\cot }^{–1}\left(\frac{\sqrt{1+\mathrm{sinx}}\mathrm{+}\sqrt{\mathrm{1}-\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}-\sqrt{\mathrm{1}-\mathrm{sinx}}}\right)\\ ={\cot }^{–1}\left(\frac{\sqrt{{\sin }^{\mathrm{2}}\frac{\mathrm{x}}{2}+{\cos }^{\mathrm{2}}\frac{\mathrm{x}}{2}+2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}\mathrm{+}\sqrt{{\sin }^{\mathrm{2}}\frac{\mathrm{x}}{2}+{\cos }^{\mathrm{2}}\frac{\mathrm{x}}{2}-2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}}{\sqrt{{\sin }^{\mathrm{2}}\frac{\mathrm{x}}{2}+{\cos }^{\mathrm{2}}\frac{\mathrm{x}}{2}+2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}-\sqrt{{\sin }^{\mathrm{2}}\frac{\mathrm{x}}{2}+{\cos }^{\mathrm{2}}\frac{\mathrm{x}}{2}-2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}}\right)\\ ={\cot }^{–1}\left(\frac{\sqrt{{(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})}^2}\mathrm{+}\sqrt{{(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})}^2}}{\sqrt{{(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})}^2}-\sqrt{{(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})}^2}}\right)\\ ={\cot }^{–1}\left(\frac{\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}+\hspace{0.17em}\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}}{\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}-(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})}\right)\\ ={\cot }^{–1}\left(\frac{2\cos \frac{\mathrm{x}}{2}}{2\sin \frac{\mathrm{x}}{2}}\right)\\ ={\cot }^{–1}\left(\cot \frac{\mathrm{x}}{2}\right)=\frac{\mathrm{x}}{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.46

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\tan }^{–1}\left(\frac{\sqrt{1+\mathrm{x}}+\sqrt{1–\mathrm{x}}}{\sqrt{1+\mathrm{x}}–\sqrt{1–\mathrm{x}}}\right)=\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\cos }^{–1}\mathrm{x},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{1}{\sqrt{2}}\le \mathrm{x}\le 1\\ [\mathrm{Hint}:\mathrm{Putx}=\cos 2\mathrm{\theta }]\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}={\tan }^{–1}\left(\frac{\sqrt{1+\mathrm{x}}+\sqrt{\mathrm{1}-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{1}-\mathrm{x}}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\sqrt{1+\cos 2\mathrm{\theta }}\mathrm{+}\sqrt{\mathrm{1}-\cos 2\mathrm{\theta }}}{\sqrt{1+\cos 2\mathrm{\theta }}-\sqrt{\mathrm{1}-\cos 2\mathrm{\theta }}}\right)[\mathrm{Putting}\mathrm{x}=\cos 2\mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\sqrt{2{\cos }^{\mathrm{2}}\mathrm{\theta }}\mathrm{+}\sqrt{2{\sin }^2\mathrm{\theta }}}{\sqrt{2{\cos }^{\mathrm{2}}\mathrm{\theta }}-\sqrt{2{\sin }^2\mathrm{\theta }}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\mathrm{cos\theta }+\mathrm{sin\theta }}{\mathrm{cos\theta }-\mathrm{sin\theta }}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{1+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{1-\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{1+\mathrm{tan\theta }}{1-\mathrm{tan\theta }}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\tan (\frac{\mathrm{\pi }}{4}-\mathrm{\theta })=\frac{\mathrm{\pi }}{4}-\mathrm{\theta }=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.47

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\sin }^{–1}\left(\frac{1}{3}\right)=\frac{9}{4}{\sin }^{–1}\left(\frac{2\sqrt{2}}{3}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{9}\mathrm{\pi }}{\mathrm{8}}-\frac{\mathrm{9}}{\mathrm{4}}{\sin }^{–1}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\\ \hspace{0.17em}\hspace{0.17em}=\frac{9}{4}\{\frac{\mathrm{\pi }}{2}-{\sin }^{–1}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\}\\ \hspace{0.17em}\hspace{0.17em}=\frac{9}{4}{\cos }^{-1}\left(\frac{1}{3}\right)...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{let}\mathrm{x}={\cos }^{-1}\left(\frac{1}{3}\right)\\ \Rightarrow \mathrm{cosx}=\frac{1}{3}\\ \Rightarrow \mathrm{sinx}=\sqrt{1-{\cos }^2\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\sqrt{1-{\left(\frac{1}{3}\right)}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{2\sqrt{2}}{3}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)\\ \therefore {\cos }^{-1}\left(\frac{1}{3}\right)={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{from}\mathrm{eq}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{9}{4}{\cos }^{-1}\left(\frac{1}{3}\right)\\ \mathrm{\hspace{0.17em}}=\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.48

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\tan }^{–1}\left(\frac{\sqrt{1+\mathrm{x}}+\sqrt{1–\mathrm{x}}}{\sqrt{1+\mathrm{x}}–\sqrt{1–\mathrm{x}}}\right)=\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\cos }^{–1}\mathrm{x},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{1}{\sqrt{2}}\le \mathrm{x}\le 1\\ [\mathrm{Hint}:\mathrm{Putx}=\cos 2\mathrm{\theta }]\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}={\tan }^{–1}\left(\frac{\sqrt{1+\mathrm{x}}+\sqrt{\mathrm{1}-\mathrm{x}}}{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{1}-\mathrm{x}}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\sqrt{1+\cos 2\mathrm{\theta }}\mathrm{+}\sqrt{\mathrm{1}-\cos 2\mathrm{\theta }}}{\sqrt{1+\cos 2\mathrm{\theta }}-\sqrt{\mathrm{1}-\cos 2\mathrm{\theta }}}\right)[\mathrm{Putting}\mathrm{x}=\cos 2\mathrm{\theta }]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\sqrt{2{\cos }^{\mathrm{2}}\mathrm{\theta }}\mathrm{+}\sqrt{2{\sin }^2\mathrm{\theta }}}{\sqrt{2{\cos }^{\mathrm{2}}\mathrm{\theta }}-\sqrt{2{\sin }^2\mathrm{\theta }}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{\mathrm{cos\theta }+\mathrm{sin\theta }}{\mathrm{cos\theta }-\mathrm{sin\theta }}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{1+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}{1-\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\left(\frac{1+\mathrm{tan\theta }}{1-\mathrm{tan\theta }}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\tan }^{–1}\tan (\frac{\mathrm{\pi }}{4}-\mathrm{\theta })=\frac{\mathrm{\pi }}{4}-\mathrm{\theta }=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.49

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{9\mathrm{\pi }}{8}-\frac{9}{4}{\sin }^{–1}\left(\frac{1}{3}\right)=\frac{9}{4}{\sin }^{–1}\left(\frac{2\sqrt{2}}{3}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{9}\mathrm{\pi }}{\mathrm{8}}-\frac{\mathrm{9}}{\mathrm{4}}{\sin }^{–1}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\\ \hspace{0.17em}\hspace{0.17em}=\frac{9}{4}\{\frac{\mathrm{\pi }}{2}-{\sin }^{–1}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\}\\ \hspace{0.17em}\hspace{0.17em}=\frac{9}{4}{\cos }^{-1}\left(\frac{1}{3}\right)...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{let}\mathrm{x}={\cos }^{-1}\left(\frac{1}{3}\right)\\ \Rightarrow \mathrm{cosx}=\frac{1}{3}\\ \Rightarrow \mathrm{sinx}=\sqrt{1-{\cos }^2\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\sqrt{1-{\left(\frac{1}{3}\right)}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{2\sqrt{2}}{3}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)\\ \therefore {\cos }^{-1}\left(\frac{1}{3}\right)={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{from}\mathrm{eq}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{9}{4}{\cos }^{-1}\left(\frac{1}{3}\right)\\ \mathrm{\hspace{0.17em}}=\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.50 Solve 2tan^-1(cosx) = tan^-1(2cosecx)

Ans.

$\begin{array}{l}{\text{2tan}}^{\text{-1}}\left(\text{cosx}\right)={\text{tan}}^{\text{-1}}\left(\text{2cosecx}\right)\\ \Rightarrow {\text{tan}}^{\text{-1}}\left(\frac{\text{2cosx}}{1-{\cos }^{2}x}\right)={\text{tan}}^{\text{-1}}\left(\frac{\text{2}}{\sin x}\right)[\because 2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)]\\ \Rightarrow \frac{\text{2cosx}}{1-{\cos }^{2}x}=\frac{\text{2}}{\sin x}\\ \Rightarrow \frac{\text{cosx}}{{\sin }^{2}x}=\frac{\text{1}}{\sin x}\\ \Rightarrow \cos x=\sin x\\ \Rightarrow 1=\tan x\\ \Rightarrow \tan \frac{\pi }{4}=\tan x\\ \therefore x=\frac{\pi }{4}\end{array}$

Q.51

$\begin{array}{l}Solve:\\ ta{n}^{–1}\frac{1-x}{1+x}=\frac{1}{2}ta{n}^{–1}x,(x>0)\end{array}$

Ans.

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\tan }^{–1}\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{2}}{\tan }^{–1}\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2{\tan }^{–1}\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}={\tan }^{–1}\mathrm{x}\\ \Rightarrow {\tan }^{–1}\left\{\frac{2\left(\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}\right)}{1-{\left(\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}\right)}^2}\right\}={\tan }^{–1}\mathrm{x}\\ \Rightarrow \mathrm{\hspace{0.17em}}\frac{2\left(\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}\right)}{1-{\left(\frac{\mathrm{1}-\mathrm{x}}{1+\mathrm{x}}\right)}^2}=\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{2(1-\mathrm{x})(1+\mathrm{x})}{{(1+\mathrm{x})}^2-{(1-\mathrm{x})}^2}=\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{2(1-{\mathrm{x}}^2)}{4\mathrm{x}}=\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}1-{\mathrm{x}}^2=2{\mathrm{x}}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1=3{\mathrm{x}}^2\\ \Rightarrow \mathrm{x}=\pm \frac{1}{\sqrt{3}}\end{array}$

Q.52

$\begin{array}{l}\sin \left({\tan }^{-1}\mathrm{x}\right),\mathrm{}\left|\mathrm{x}\right|\mathrm{\hspace{0.33em}}<\mathrm{\hspace{0.33em}}1\hspace{0.33em}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{a}\right)\mathrm{\hspace{0.33em}}\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\\ \left(\mathrm{b}\right)\mathrm{\hspace{0.33em}}\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\\ \left(\mathrm{c}\right)\mathrm{\hspace{0.33em}}\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\\ \left(\mathrm{d}\right)\mathrm{\hspace{0.33em}}\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\end{array}$

Ans.

\begin{array}{l}\text{sin}\left({\text{tan}}^{\text{-1}}\text{x}\right)\\ Let{\text{tan}}^{\text{-1}}\text{x}=\theta \Rightarrow x=\tan \theta \\ \Rightarrow {\sec }^2\theta =1+{\tan }^2\theta \\ \Rightarrow \frac{1}{{\cos }^2\theta }=1+x^2\\ \Rightarrow {\cos }^2\theta =\frac{1}{1+x^2}\\ Since,\\ \Rightarrow \sin \theta =\sqrt{1-{\cos }^2\theta }\\ =\sqrt{1-\frac{1}{1+x^2}}\\ =\sqrt{\frac{1+x^2-1}{1+x^2}}=\frac{x}{\sqrt{1+x^2}}\\ \theta ={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\\ \Rightarrow {\text{tan}}^{\text{-1}}\text{x}={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\\ Therefore,\\ \text{sin}\left({\text{tan}}^{\text{-1}}\text{x}\right)=\sin \left\{{\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\right\}\\ =\frac{x}{\sqrt{1+x^2}}\\ Thus,\text{option ‘D’ is correct solution}\text{.}\end{array}

Q.53

$\begin{array}{l}{\sin }^{–1}(1-\mathrm{x})-2{\sin }^{–1}\mathrm{x}=\frac{\mathrm{\pi }}{2},\mathrm{then}\mathrm{x}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)0,\frac{1}{2}\\ (\mathrm{B}1,\frac{1}{2}\\ \left(\mathrm{C}\right)0\\ \left(\mathrm{D}\right)\frac{1}{2}\end{array}$

Ans.

\begin{array}{l}{\text{sin}}^{\text{-1}}\left(\text{1}-\text{x}\right)-{\text{2sin}}^{\text{-1}}\text{x}=\frac{\pi }{2}\\ 1-x=\sin \left(\frac{\pi }{2}+2{\sin }^{-1}x\right)\\ =\cos \left(2{\sin }^{-1}x\right)\left[\because sin\left(\frac{\pi }{2}+x\right)=\cos x\right]\\ =\cos \left\{{\cos }^{-1}\left(1-2x^2\right)\right\}\left(\begin{array}{l}Let{\sin }^{-1}x=\theta \text{and cos2}\theta \text{=1}-{\text{2sin}}^{\text{2}}\text{x}\\ \text{2}\theta ={\cos }^{-1}\left(1-2{\sin }^{2}x\right)\\ \Rightarrow 2{\sin }^{-1}x={\cos }^{-1}\left(1-x^2\right)\end{array}\right)\\ \Rightarrow 1-x=1-2x^2\\ \Rightarrow 2x^2-x=0\\ \Rightarrow x\left(2x-1\right)=0\\ \Rightarrow x=0\text{or 2x}-\text{1}=\text{0}\\ \Rightarrow x=0\text{or x}=\frac{1}{2}\\ \text{But on substitution of x}=\frac{1}{2}\\ L.H.S.={\text{sin}}^{\text{-1}}\left(\text{1-x}\right)-{\text{2sin}}^{\text{-1}}\text{x}\\ ={\text{sin}}^{-\text{1}}\left(\text{1}-\frac{\text{1}}{2}\right)-{\text{2sin}}^{\text{-1}}\left(\frac{\text{1}}{2}\right)\\ =\frac{\pi }{6}-2\left(\frac{\pi }{6}\right)\\ =-\frac{\pi }{6}\ne \frac{\pi }{2}\ne R.H.S.\\ \therefore x=\frac{1}{2}\text{is not the solution of given equation}\text{.}\\ \text{Thus, x}=\text{0 is the solution of given equation}\text{.}\\ \text{Therefore, option ‘C’ is correction solution of given equation}\text{.}\end{array}

Q.54

$\begin{array}{l}{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-{\tan }^{-1}\left(\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x}+\mathrm{y}}\right)\mathrm{}\mathrm{is}\mathrm{equal}\mathrm{to}\\ \left(\mathrm{A}\right)\frac{\mathrm{\pi }}{2}\\ \left(\mathrm{B}\right)\frac{\mathrm{\pi }}{3}\\ \left(\mathrm{C}\right)\mathrm{\hspace{0.33em}}\frac{\mathrm{\pi }}{4}\\ \left(\mathrm{D}\right)\mathrm{\hspace{0.33em}}\frac{-3\mathrm{\pi }}{4}\end{array}$

Ans.

$\begin{array}{l}{\tan }^{–1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-{\tan }^{–1}\left(\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x}+\mathrm{y}}\right)\\ ={\tan }^{–1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-{\tan }^{–1}\left(\frac{\frac{\mathrm{x}}{\mathrm{y}}-\mathrm{1}}{\frac{\mathrm{x}}{\mathrm{y}}+1}\right)\\ ={\tan }^{–1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-{\tan }^{–1}\left\{\frac{-(1-\frac{\mathrm{x}}{\mathrm{y}})}{(1+\frac{\mathrm{x}}{\mathrm{y}})}\right\}\\ ={\tan }^{–1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+{\tan }^{–1}\left\{\frac{(1-\frac{\mathrm{x}}{\mathrm{y}})}{(1+\frac{\mathrm{x}}{\mathrm{y}})}\right\}[\because {\tan }^{-1}(-\mathrm{x})=-{\tan }^{-1}\mathrm{x}]\\ ={\tan }^{–1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+{\tan }^{–1}\left(1\right)-{\tan }^{–1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\\ ={\tan }^{–1}\left(1\right)\\ =\frac{\mathrm{\pi }}{4}\\ \mathrm{Therefore},\mathrm{option}‘\mathrm{C}‘\mathrm{is}\mathrm{correct}\mathrm{solution}\mathrm{.}\end{array}$